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I know that the action, in Classical Mechanics, is a functional of the path of a physical system, such that

"the path actually followed by a physical system is that for which the action is minimized, or more generally, is stationary" (Wikipedia)

and this definition can be extended to the Path Integral formulation of Quantum Mechanics and Quantum Field Theory. This is how I've always used the action: I want to find the path, I have my Lagrangian, I integrate it and minimize it, thus finding the path, not caring about the value of the minimized action.
My question is: is the value of the action at the said minimum (or stationary point) ever important? Maybe it isn't for CM, but it is for QM or QFT? Does it matter if the action is Minkowskian or Euclidean?

My doubt comes from trying to understand Coleman's Fate of the false vacuum, which uses the fact that the decay rate of the metastable vacuum $\Gamma$ is $$\Gamma\propto e^{-S_E/\hbar}\tag{2.2 and 2.18}$$ where $S_E$ is the Euclidean action: for this case, the value of the action seems to be rather important.

Addendum: my initial question was about the sign of the Euclidean action of the nucleation rate mentioned above: if the sign is "allowed to be negative", then the nucleation can go from being exponentially suppressed to exponentially favored, which I don't think is physically meaningful. Later I thought that since I've never used the value of the action at its minimum, maybe this value isn't important, and therefore neither is its sign (if, for example, I can shift a positive minimum enough to make it negative, and vice versa).

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  • $\begingroup$ In the Euclidean picture it's clear that the value of the action is how much 'weight' a given field configuration has. This for example allows you to estimate the sizes of instanton effects from the value of the action (or at the very least, how such effects scale when you change parameters); I'm pretty sure about this but let me dig up some references $\endgroup$ Jul 8, 2022 at 14:58
  • $\begingroup$ Perhaps the confusion is that the stationary path occurs where the action takes its minimum value. So it matters that the value be at its minimum, whatever that turns out to be. $\endgroup$
    – mmesser314
    Jul 8, 2022 at 14:58
  • $\begingroup$ @JoshuaLin References would be highly appreciated, thanks. So you're saying that the value of the action is only important in the Euclidean picture? If so, how could a Wick-rotation change so much about the information that one can "grab" from a functional? $\endgroup$ Jul 8, 2022 at 15:01
  • $\begingroup$ @mmesser314 if that is the case, then there's no confusion at all: I'm aware that the stationary math occurs where the action is minimized - I'm just not sure of the "whatever that turns out to be", hence the question. Could you explain why the specific value doesn't matter? $\endgroup$ Jul 8, 2022 at 15:04
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    $\begingroup$ I didn't mean anything mysterious. Take an action. Find the stationary path. Add 5 to the action. You get the same stationary path. The minimum value has changed, but the new minimum still occurs in the same place. $\endgroup$
    – mmesser314
    Jul 8, 2022 at 15:08

4 Answers 4

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  1. For the physical meaning of the action in classical physics, see e.g. this & this related Phys.SE posts.

  2. In the quantum path integral the action is important in calculating probability amplitudes.

    An overall phase factor is not physical relevant, but relative phase factors are.

    See in particular the semiclassical WKB/stationary phase approximation, cf. e.g. this Phys.SE post.

    In Coleman's example the Euclidean action $$S_E~=~2\int_{\vec{q}+0}^{\vec{\sigma}} ds \sqrt{2V}\tag{2.5+2.18}$$ is the action of a single instanton (which he calls a bounce), cf. e.g. this related Phys.SE post. Coleman's calculation is normalized relative to the 0-bounce sector. The value of the action obviously affects the decay rate (2.2). The fact that the Euclidean action enters can be viewed as an effect of a complex integration contour in the method of steepest descent.

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  • $\begingroup$ So the action is classically used only as a functional to be minimized, and so classical physics is invariant if $S\rightarrow S+5$, while the quantum path integral uses the value of the action to compute probability amplitudes, that therefore are not invariant if $S\rightarrow S+5$? It's the quantumness that seals the deal, not the Euclideaness, right? $\endgroup$ Jul 8, 2022 at 15:40
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    $\begingroup$ An overall shift of the action in the path integral or the statistical partition function is not physical relevant as well. $\endgroup$
    – Qmechanic
    Jul 8, 2022 at 19:23
  • $\begingroup$ this changes a lot of what I thought I understood. How is Coleman's Euclidean action different from the action in the path integral? It must be different, since a shift of Coleman's action is relevant, while a shift of the path integral's action isn't. Is it the fact that in the path integral I integrate the action, but I don't integrate the content of Coleman's one? If that's the case, I'd be more inclined to say that it's the integration that's invariant under overall shifts, not the action itself, therefore I suspect this isn't the correct reasoning. $\endgroup$ Jul 9, 2022 at 0:00
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jul 9, 2022 at 8:54
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    $\begingroup$ Coleman's action $S_E$ is relative, i.e. a difference between the absolute action of different instanton sectors. $\endgroup$
    – Qmechanic
    Jul 9, 2022 at 17:51
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The answer is slightly different in classical and quantum mechanics.

In classical mechanics, two Lagrangians are physically equivalent if one is a linear rescaling of another: $\mathcal{L}_1 \equiv a \mathcal{L}_2 + b$, where $a$ and $b$ are real constants. Neither the overall prefactor nor the constant offset have any physical significance. That's because the extremal path isn't affected by either parameter.

In quantum mechanics, two Lagrangians are physically equivalent if they differ by a constant: $\mathcal{L}_1 \equiv \mathcal{L}_2 + b$. However, the overall prefactor does have physically measurable consequences. (It controls how much off-shell paths contribute to the path integral, or, very loosely, how important the "quantum fluctuations" are - although I often find that term to be more misleading than clarifying.)

In both cases, the numerical value of the action at the extremal path has no physical significance.

You can think of the Lagrangian as just being a potential on a rather abstract space (the space of paths or field configurations, not physical space). Like any other potential, only differences between its values at different "points" are physically important, not its absolute value (i.e. the action) at any single "point". So you're always free to shift it by an overall additive constant.

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The value at a given point is never important, only the stationary path. Indeed, we find:

Any function which generates the correct equations of motion, in agreement with physical laws, can be taken as a Lagrangian.

It can be shown that there are mathematical transformations which change the value of the action but do not change the path. In particular, you can scale and add to the Lagrangian without changing the results ($L^\prime = aL+b$). Even if you fix the endpoints of your path, you can still transform it with: $$L'(q,\dot q,t) = L(q,\dot q,t) + \frac{\mathrm{d}f(q,t)}{\mathrm{d}t}$$

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    $\begingroup$ Is this true regardless of the "quantumness" of the phenomenon (CM or QM) and the metric used in the action (Minkowskian or Euclidean)? If so, am I mistaken in interpreting eq. (2.18) of the linked paper as "the exponential of the negative of the value of the action" (since its value isn't important, ever)? $\endgroup$ Jul 8, 2022 at 14:55
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Indeed in Classical Mechanics the value of Hamilton's action is not a factor; it is the derivative of Hamilton's action that is physically significant.


Arbitrary choice of zero point

Hamilton's action: $E_k - E_p$

As we know: potential energy does not have an intrinsic zero point. The quantity that enters calculation is difference of potential.

(Of course, in many cases one particular choice of zero point of potential energy is most convenient for the purpose of calculation, but in any case: the quantity that enters calculation is difference of potential, accommodating that the choice of zero point is arbitrary.)

As we know: the amount of kinetic energy that is attributed to an object is kinetic energy relative to some reference. In that sense kinetic energy does not have an intrinsic zero point; we can transform from one inertial coordinate system to another, changing the kinetic energy that is attributed, and there is no physical significance in that change.

Conclusion: Hamilton's action does not have an intrinsic zero point because neither of the constituent parts of Hamilton's action have an intrinsic zero point.


Position, velocity, acceleration

We have that the derivative of Hamilton's action is physically significant, while at the same time the value of Hamilton's action does not have a physical interpretation. It's interesting to see how that comes about.

In theory of motion (classical mechanics) we have the trio: position, velocity, acceleration. Position and velocity are relative, acceleration is absolute.

In mathematics we are accustomed to thinking of the process of taking a derivative as an operation that moves from upstream to downstream. In mathematics a derivative is thought of as a quantity that inherits all of its properties from the quantity that it is derived from.

In physics: the quantity that acceleration is the derivative of (velocity) is relative, whereas acceleration is absolute.


Energy

In the case of energy there is a parallel with how position, velocity, and acceleration relate to each other.

The quantity potential energy is obtained by evaluating the integral of force with respect to spatial displacement. (And the quantity kinetic energy is obtained by evaluating the integral of $ma$ (mass times acceleration))

To evaluate energy in calculation: we take the derivative of the energy with respect to the spatial coordinate (the reverse operation of how the energy is obtained).

Specifically: taking the derivative of kinetic energy with respect to the spatial coordinate results in $ma$ (mass times acceleration)

$$ \frac{d(\tfrac{1}{2}mv^2)}{ds} = \tfrac{1}{2}m\left( 2v\frac{dv}{ds} \right) = m\frac{ds}{dt}\frac{dv}{ds} = m\frac{dv}{dt} = ma $$

Circling back to the start of this answer: the value of Hamilton's action is not a factor; it is the derivative of Hamilton's action that is physically significant. (The derivative of Hamilton's action with respect to the spatial coordinate)

Taking the derivative of the energy is an operation that parallels the operation of moving from velocity to acceleration.

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